How Does Sun's Luminosity Affect E Field Amplitude at Earth's Atmosphere?

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The luminosity of the sun, at 3.9E26 watts, influences the irradiance experienced at Earth's atmosphere, which is calculated using the inverse square law. At a distance of 1.5E11 m, the irradiance is determined to be 8.78E-3 watts/m^2. The amplitude of the electric field can then be calculated using the formula E = sqrt(I/ε), where ε is the permittivity of air, approximately 8.85E-12 F/m. This results in an electric field amplitude of approximately 1.05E-5 volts/m at that distance. Thus, while luminosity does not directly affect electric field amplitude, it significantly impacts the irradiance that leads to the amplitude calculation.
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I'm trying to determine how the luminosity affects the amplitude of the resulting E field at a given distance. I know that the irradiance and amplitude are linked by the fact that the irradiance is proportional to the square of the amplitude of the E field but I don't have that.
I have luminosity of the sun 3.9E26W and want the amplitude of the electric field at a distance of 1.5e11 m away (the top of the Earth's atmosphere).
Anyone give me any hints?
 
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they're not related, where did your question came from? i guess you misunderstand the question
 


The luminosity of a source, such as the sun, refers to the total amount of energy emitted per unit time. In the case of the sun, this is equal to 3.9E26 watts. The irradiance, on the other hand, refers to the amount of energy per unit area received by a surface. This is related to the amplitude of the electric field by the equation E = sqrt(I/ε), where E is the amplitude of the electric field, I is the irradiance, and ε is the permittivity of the medium.

To determine the amplitude of the electric field at a distance of 1.5E11 m away from the sun, we can use the inverse square law, which states that the irradiance is inversely proportional to the square of the distance from the source. This means that the irradiance at a distance of 1.5E11 m will be 1/((1.5E11)^2) = 4.44E-22 times the irradiance at the surface of the sun.

Using the value of the sun's luminosity and the inverse square law, we can calculate the irradiance at the top of the Earth's atmosphere, which is equal to 8.78E-3 watts/m^2. Plugging this value into the equation for electric field amplitude, we get E = sqrt(8.78E-3/ε). The permittivity of air at standard temperature and pressure is approximately 8.85E-12 F/m, so the amplitude of the electric field at a distance of 1.5E11 m away from the sun is approximately 1.05E-5 volts/m.

In summary, the luminosity of a source does not directly affect the amplitude of the electric field at a given distance. However, it does affect the irradiance, which in turn affects the amplitude of the electric field. By using the inverse square law and the equation for electric field amplitude, we can determine the amplitude of the electric field at a given distance from a source with a known luminosity.
 
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